Question

Calcule o limite abaixo:

lim 2 - √x+1 / x² - 9

X tende a 3

Answer 1

$\lim_{ x \to 3} \frac{2-\sqrt{x+1}}{x^2-9} =\lim_{ x \to 3} \frac{2-\sqrt{x+1}}{x^2-3^2} \\\\ \text{diferenca dos quadrados} (a^2-b^2)=(a+b)(a-b)\\\\ \lim_{ x \to 3} \frac{2-\sqrt{x+1}}{(x-3)(x+3)}\\\\ \text{multiplica pelo conjugado do numerador} \\\\ \lim_{ x \to 3} \frac{2-\sqrt{x+1}}{(x-3)(x+3)} * \frac{2+\sqrt{x+1}}{2+\sqrt{x+1}} \\\\ \lim_{ x \to 3} \frac{(2-\sqrt{x+1})(2+\sqrt{x+1})}{(x-3)(x+3)(2+\sqrt{x+1})}\\\\ \lim_{ x \to 3} \frac{2^2-(\sqrt{x+1})^2}{(x-3)(x+3)(2+\sqrt{x+1})}$

$\lim_{ x \to 3} \frac{4-(x+1)}{(x-3)(x+3)(2+\sqrt{x+1})}\\\\ \lim_{ x \to 3} \frac{3-x}{(x-3)(x+3)(2+\sqrt{x+1})}\\\\ \lim_{ x \to 3} \frac{-(x-3)}{(x-3)(x+3)(2+\sqrt{x+1})}\\\\ \lim_{ x \to 3} \frac{-1}{(x+3)(2+\sqrt{x+1})}= \frac{-1}{(3+3)(2+\sqrt{3+1})} = \frac{-1}{24}$